How to prove $A\cap (B\setminus C) = (A\cap B)\setminus(A\cap C) = (A\cap B)\setminus C$? I do understand how to prove something like 
$$A\cap (B\setminus C)= (A\cap B)\setminus C$$ but I don't really understand how do I do it with something like this $$A\cap (B\setminus C) = (A\cap B)\setminus(A\cap C) = (A\cap B)\setminus C$$
 A: In fact $X\setminus Y=X\cap Y^\complement$
So the above equality is just the expression of intersection commutativity and associativity.
$A\cap(B\setminus C)=A\cap(B \cap C^\complement)=(A\cap B)\cap C^\complement=(A\cap B)\setminus C$
The middle statement is less immediate, it is easier to develop it.
We will use the rule : $(X\cap Y)^\complement=X^\complement \cup Y^{\complement}$
You get $(A\cap B)\setminus(A\cap C)=(A\cap B)\cap(A^\complement\cup C^\complement)$
And now we distribute $\cup$ over $\cap$ 
Same as you do with addition an,d multiplication. eg. $(a\times b)\times(a'+c')=(aba')+(abc')$
$\cdots=\underbrace{(A\cap B\cap A^\complement)}_{\varnothing}\cup(A\cap B\cap C^\complement)=(A\cap B)\setminus C$
The first term being empty set since $(A\cap A^\complement=\varnothing)$
Here is a table summarizing De Morgan's laws

A: The way to solve questions about equality of subsets is to appeal to the definition, and show that each is a subset of the other (as Andrew Li has pointed out). In your case, you need to prove that three things are equal to each other. One way to do this is to show that the first two things are equal to each other, and then that the second two things are equal to each other, which by the transitivity of equality will imply that the first and third things are equal (since they are both equal to the second thing.
In your case, you've already proved that $A \cap (B \setminus C) = (A \cap B) \setminus C$ (or at least you know how to prove this), so that the first and third sets are equal. Now you only need to show that the first and second sets are equal, or that the second and third sets are equal, which will imply that all three sets are equal. 
Of those two options, I think it is easier to show that the second and third sets are equal, and you can do this (again, as Andrew Li suggested), by showing that each set is a subset of the other. 
